Question

Mass of Electron = 9.10939 x 10−31kg An electron remains suspended between the surface of the Earth (assumed neutral) and a fixed positive point charge, at a distance of 11.62 m from the point charge. Determine the charge required for this to happen. The acceleration due to gravity is 9.8 m/s 2 and the Coulomb constant is 8.98755 × 109 N · m2 /C 2 . Answer in units of C.

Answer 1

`8.4\cdot 10^(-19)C`

Step-by-step explanation:

The electron remains suspended is the electric force acting on it (upward) is equal in magnitude to the force of gravity (downward) acting on it.

The electric force on the electron is given by:

`F_E=(keq)/(r^2)`

where:

k is the Coulomb's constant

`e=1.6\cdot 10^(-19)C` is the magnitude of the electric charge

q is the unknown positive charge

r = 11.62 m is the distance between the electron and the charge

The force of gravity on the electron is

`F_G=mg`

where

`m=9.11\cdot 10^(-31) kg` is the mass of the electron

`g=9.8 m/s^2` is the acceleration due to gravity

Equating the two forces and solving for q, we find:

`(keq)/(r^2)=mg\\q=(mgr^2)/(ke)=((9.11\cdot 10^(-31))(9.8)(11.62)^2)/((9\cdot 10^9)(1.6\cdot 10^(-19)))=8.4\cdot 10^(-19)C`